Answer:
26 lbf
Explanation:
The mass of the satellite is the same regardless of where it is.
The weight however, depends on the acceleration of gravity.
The universal gravitation equation:
g = G * M / d^2
Where
G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))
M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)
d: distance to that body
15000 miles = 24140 km
The distance is to the center of Earth.
Earth radius = 6371 km
Then:
d = 24140 + 6371 = 30511 km
g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2
Then we calculate the weight:
w = m * a
w = 270 * 0.43 = 116 N
116 N is 26 lbf
Answer:
Metal wireways are sheet metal "U"s with removable housing for protecting electrical equipment, wires, and cables.
Explanation:
These are especially used to run wire in manufacturing environments.
Answer:
The correct answer is option 'B': Load is far from fulcrum and the effort is applied near the fulcrum
Explanation:
A lever works on the principle of balancing of torques. The torque about the fulcrum by the load should be equal to the torque by the applied effort. Since we know that the torque is proportional to both the force and the distance it is applied from the distance from the axis of rotation. A lever is used when we need to lift a heavy load by utilizing this effect of the lever arm.
A mechanical disadvantage occurs when we are not able to lift the weight easily due to the fact we apply effort near the fulcrum.
Answer:
3A
Explanation:
Using Ohms law U=I×R solve for I by I=U/R
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.